Divergent Countable Sum from Uncountable Sum

Question

While pondering on the counting measure recently, I considered the following:

Let us define $\sum_{x \in X}f(x)$ as $\int_X f(x) d\mu$ where $\mu$ is the counting measure

Suppose $\sum_{x \in X}f(x) = \infty$ where $X$ is uncountable and $0 \le f \le \infty$

Does there exist some countable subset $S \subset X$ such that $\sum_{x \in S}f(x) = \infty$?

All the examples I tried worked out, but I'm not sure about the validity of the result in general - in fact, I remain quite skeptical. I imagine there is a simple proof either way, but haven't thought of one.

Does anyone know such a proof?

Answer 1

Since $f \ge 0$ then either $f>0$ for only countably many terms, or $f>0$ for uncountably many terms. Then uncountably many must be bigger than some $1/n$ for some $n$. Hence we have the result either way.

Answer 2

Same result as David's but expressed in a more complicated way.

If $f\cdot 1_X$ is $|\cdot|$ measurable there is a sequence of simple functions $s_n$ such that $s_n \le f \cdot 1_X$ and $\int s_n \to \infty$.

If the support of all the $s_n$ is countable, then the union of the supports is countable and the integral over this set is $\infty$.

If the support of any of the $s_n$ is uncountable, then we have $f \ge s_n \ge \alpha \cdot1_A$ for some uncountable $A$ with $\alpha >0$. Now choose any countable subset of $A$ to get the desired result.